In this entry we compute the derivative of the hyperbolic functions $\sinh(x)$ and $\cosh(x)$ .

Recall that by definition:

\begin{eqnarray*} \sinh(x)&:=&\frac{e^x-e^{-x}}{2}\\ \cosh(x)&:=&\frac{e^x+e^{-x}}{2}. \end{eqnarray*} Therefore:

\begin{eqnarray*} \frac{d}{dx}\sinh(x) &=& \frac{d}{dx}\left(\frac{e^x-e^{-x}}{2}\right)\\ &=& \frac{1}{2}\cdot\frac{d}{dx}\left(e^x-e^{-x}\right)\\ &=& \frac{1}{2}\cdot\left(e^x-(-e^{-x})\right)\\ &=& \frac{e^x+e^{-x}}{2}\\ &=& \cosh(x). \end{eqnarray*} Similarly $\displaystyle \frac{d}{dx}\cosh(x)=\sinh(x)$ . Using the quotient rule, we compute the derivative of $\displaystyle \tanh(x)=\frac{\sinh(x)}{\cosh(x)}$ : $$\frac{d}{dx}\tanh(x)=\frac{\cosh^2(x)-\sinh^2(x)}{\cosh^2(x)}=\frac{1}{\cosh^2(x)}$$ where we have used the equality $\cosh^2(x)-\sinh^2(x)=1$ .